To reduce the differential equation $\frac{dy}{dx} + P(x)y = Q(x)y^n$ to the linear form,the substitution is

Let $y=y(x)$ be the solution curve of the differential equation $\sec y \frac{dy}{dx} + 2x \sin y = x^3 \cos y$,with the initial condition $y(1) = 0$. Then the value of $y(\sqrt{3})$ is equal to:

Let $y=y(x)$ be the solution of the differential equation $x \log _e x \frac{d y}{d x}+y=x^2 \log _e x, (x > 1)$. If $y(2)=2$,then $y(e)$ is equal to

Let $f: R \to R$ be such that $f(xy) = f(x)f(y)$ for all $x, y \in R$ and $f(0) \ne 0$. Let $g: [1, \infty) \to R$ be a differentiable function such that $x^2 g(x) = \int_1^x (t^2 f(t) - t g(t)) dt$. Then $g(2)$ is equal to:

$A$ function $y = f(x)$ satisfies $(x + 1)f'(x) - 2(x^2 + x)f(x) = \frac{e^{x^2}}{(x + 1)}$. If $f(0) = 5$,then $f(x)$ is:

The solution of $(1+y^2)+(x-e^{\tan ^{-1} y}) \frac{dy}{dx}=0$ is